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Chris and Evan are outstanding runners. One day in practice they do a 1500 meter training run. Being competitors, Chris and Evan can’t do any training run without it turning into a competition, so they decide to race. Chris and Evan were dead even until Chris tripped at the 1100 meter mark. His misfortune allowed Evan to take a 15 meter lead. Jolted by adrenaline Chris immediately ups his pace to 6.7 meters/second. Evan continues at his consistent pace of 6.4 meters/second. Who wins the race?

This problem mimics some themes we often see in math problems in content, although not exactly in form. The common thread is that you have two rates. Dealing with two rates simultaneously, however, can be tricky, so it’s easier to deal only with the difference between the rates. In this case (at least at first), let’s not worry about 6.4 and 6.7. All we need to know is that Chris gains 0.3 meter every second. With a difference of 15 meters, it’s quick to figure out that it will take 15/0.3 = 50 seconds to close the gap.

Only one question remains: When Chris does catch up, will it be before or after the finish line? Since we know the two runners will be tied 50 seconds after the trip, all we need to do is multiply one of the runners’ rates by 50 seconds to figure out where on the course he will be when that occurs. Let’s chooses Chris. 6.4 m/s * 50 seconds = 320 meters. Adding that to the 1115 meters he has already traveled (since he’s already gained the 15 meter lead by the time both men are running again) and we find that Chris will be 1435 meters into the race when Evan pulls even with him.

Despite his trip, Evan will pull even with Chris and pass him down the back stretch to win the 1500 meter race.

Christian has 13 coins in his pocket. The total value of the change is $1.37. The change may be composed of pennies, nickels, dimes and quarters. How many dimes could be in Christian’s pocket?

A) 0

B) 3

C) 5

D) 7

E) 9

There are several ways to approach this problem. Many of them are wrong. That’s because the core problem (13 coins, $1.37) can be solves several different ways. If you approached this and just decided to solve it you might find two or three solutions before you found one that was in the answer choices. That’s the first important step. When you’re given a problem without a clear, obvious and quick solution look to the answer choices.

Once, we’ve decided that we’re going to work through the answer choices, there is still some uncertainty. Look at answer choice B for instance. Even if we know that we have 3 dimes, that still doesn’t sort out how many pennies, nickels and quarters we have. That brings us to tip number two: when you’re given complex information, look for opportunities to simplify. The best example of that in this case is the pennies. Because our total needs to end in a 7, and all the other coins can only get us multiples of 5, we know that we must have either 2 or 7 pennies in order to make this work. That makes quick work of checking the possibilities for answer choice B. If we have 3 dimes and 2 pennies, can we make $1.05 out of 8 coins? Well, quarters can only get us to multiples of 25, so we must have at least one nickel so we can simplify again. Can we make $1.00 composed of quarters and nickels? Nope. Doesn’t work. Test 3 dimes and 7 pennies and you’ll very quickly see that you can’t make $1.00 out of 3 of the coins we’re given.

The final tip is that challenging problems don’t need to be tough. If you went into this and said I’m not sure how to solve this, but I can work from the answer choices and I can simplify problems, you might have started by testing answer A.

If I don’t have any dimes, I might have exactly 2 pennies. If that’s the case I have 11 coins to make $1.35. I need at least 2 nickels, so now I have 9 coins to make $1.25. 4 quarters and 5 nickels would work. Done. Answer choice A is correct. (You can also make this work with 0 dimes and 7 pennies).

When you don’t know where to start, look at the answer choices and look to simplify.

One of the books I’m currently reading is Priceless: The Myth of Fair Value (and How to Take Advantage of It) by William Poundstone. It’s an interesting look as some of the psychology behind how we think about prices and perceive value. Poundstone recounts one experiment done by Kahneman and Tversky called “Linda the Bank Teller” that I’d like you to think about.

“Linda is 31 years old, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations.”

Which of the following is more likely to be true?

Linda is a bank teller.

Linda is a bank teller and is active in the feminist movement.

Astonishingly, 85% of the college students tested responded that the second statement is more likely to be true than the first, which is absolutely ridiculous. The point that Kahneman and Tversky were making is that we rely on heuristics or “rules of thumb” more often than is logically sound. The point I’d like to make is about drawing diagrams.

The information you’re given in this situation, although different in format from what you’re going to see on a standardized test is nonetheless similar. You’re given a few different pieces of information that fit together in some way that may not be intuitively obvious. One of the best ways to deal with that situation is to draw a graphical representation of what you’ve been told. When given two overlapping sets, you should draw a Venn diagram.

Now, let’s consider the two situations. First, the chance that Linda is a bank teller is represented by the region in red.

The chance that Linda is a bank teller and is active in the feminist movement is represented by the region in blue.

As you can see, the second set is a subset of the first. A drawing with only the two regions we’re concerned with would look something like this:

Linda can’t be a feminist and bank teller without being a bank teller. In this graphic form it’s pretty obvious, and yet 85% of the students missed this fact the first time around. This is a great example of the benefits of converting words into graphical forms so that you can make sure you have a good understanding of what you’re doing before you go forward with a problem.

Augie is blind. Thanks to his organized systems he is able to live alone. However, one day he finds that his laundry service has forgotten to pair his socks together. Augie has 12 black socks, 8 white socks, 6 blue socks and 4 red socks. Augie puts two random socks on.

Now, let’s ask a probability question.

What is the probability that Augie has put on a matching pair of socks?

This question tests both OR and AND probabilities. Let’s lay it out systematically.

There are four ways Augie could be wearing a matching pair of socks (black, black), (white, white), (blue, blue) or (red,red).

Since any one of these situations satisfies our condition of matching socks we want to add their individual probabilities together.

Now, we need to find those four probabilities individually. In order to pull a matching pair of black socks. In order to do that the first sock we pull out must be black. The odds of that are 12/30 (number of black socks/number of total socks) (12/30 reduces to 2/5 but we’ll save reducing for the end). But in order to pull a pair of black socks we need to pull a black sock first AND pull a black sock second. In order to pull the second black sock we’ll have to pull one of the remaining 11 black socks out of the remaining 29. Since we need both events to occur in order to get a black pair, we’ll multiply the individual probabilities to get the probability of getting a black pair:

12/30 * 11/29 = 132/870

Following the same process for the other colors gets you:

P(matching white) = 8/30 * 7/29 = 56/870

P(matching blue) = 6/30 * 5/29 = 30/870

P(matching red) = 4/30 * 3/29 = 12/870

Adding the probabilities gets a total of 230/870 = 23/87 or about 26.4%.

In this case you need to remember that when there are two individual events that must BOTH occur, you multiply their probabilities together. When there are multiple desired outcomes that are EACH satisfactory you add their probabilities.